# EM Dis Ch 1 Part 1

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Chapter 1
Vector Algebra
Part One
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Vector Analysis
Electrostatic
fields
Magnetostatic
fields
Electromagnetic
fields(wave)
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Unit Vector of a vector A
is a vector whose magnitude is
unity and its direction is along A
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Example
___
A  3ax  4ay  5az
A
aA 
|A |
Ax
| A |

3
2
2
 Ay  Az
4 5
2
2
2

2

50  7.071
3ax  4ay  5az
aA 
7.071
aA  0.42 ax  0.5657 ay  0.7071 az
| aA | 1 , , magnitude  1, , direction along A
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Position Vector of point P
is a vector directed from origin to P
e.g.
OP=P-O= 2 ax + 3 ay + 4 az
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Distance Vector
directed from one point to another
point .
e.g.
PQ=Q-P=
=(0-1)ax+(3-2)ay+(1-5)az
=-ax + ay - 4 az
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P1(2,4,4) , P2(-3,2,2)
Find the unit vector along P1P2
P1P 2  P 2  P1  (3  2)ax  ( 2  4) ay  (2  4)az
 5ax  2ay  2az
P1P 2
 5ax  2ay  2az
a P1P 2 

 0.87ax  0.348ay  0.348az
| P1P 2 |
25  4  4
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A=ax+3az
B=5ax+2ay-6az
Find:
(a) |A+B|
| A  B || 6ax  2ay  3az |
 36  4  9  7
(c) The component of A along ay
zero
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(d) Unit vector parallel to 3A+B
3 A  B  8ax  2ay  3az
| 3 A  B | 64  4  9  77
a3 A B
8ax  2ay  3az

 0.91168ax  0.22799ay  0.3419az
77
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Points P(1,-3,5) , Q(2,4,6) , R(0,3,8)
Find:
(a) Position vectors for P and R
OP  P  O  ax  3ay  5az
OR  R  O  3ay  8az
(b) Distance vector QR
QR  R  Q  (0  2)ax  (3  4)ay  (8  6)az
 2ax  ay  2az
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Points P(1,-3,5) , Q(2,4,6) , R(0,3,8)
Find:
(c) Distance between Q and R
QR  2ax  ay  2az
| QR |
4 1 4  3
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Vector Multiplication
(1) Dot Product (Scalar)
A=Ax.ax+Ay.ay+Az.az
B=Bx.ax+By.ay+Bz.az
A.B=|A| |B| cos θ
A.B=AxBx+AyBy+AzBz
Note:
A.B=-|A| |B| cos θ

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Notes:
(1) If A.B=0 θAB =90 Orthogonal
ax.ay=0
ax.az=0
ay.az=0
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(2) ax.ax=|ax| =1
2
ay.ay=|ay| =1
2
az.az=|az| =1
(3)
2
A.A=|A| |A| cos 0 =|A|
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(2) Cross Product (Vector)
A=Ax.ax+Ay.ay+Az.az
B=Bx.ax+By.ay+Bz.az
AxB=|A| |B| sin θ .an
AxB=
• Cross product is a vector
direction: Orthogonal to A and B plane
magnitude: area of parallelogram
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‫متوازي االضالع‬
Notes:
(1) AxB=-(BxA)
(2) ax x ay = az
ay x az = ax
az x ax = ay
ax x az = -ay
(1) AxA=|A| |A| sin 0 = 0
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‫‪e.g.‬‬
‫‪A=2ax-ay-2az‬‬
‫‪B=4ax+3ay+2az‬‬
‫)‪(AxB).A=|AxB||A|cos 90= 0 (normal‬‬
‫)‪(AxB).B=0 (normal‬‬
‫العمودً علي المستوى عمودً علي أً متجه يحتويه هذا المستوى‬
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A=ax+3az
B=5ax+2ay-6az
Find the angle between vector A and B
A.B=5+0-18=-13
| A | 1  9  10
| B | 25  4  36  65
A.B | A || B | cos
 13
cos 
 0.5099
10 65
  cos1 (0.5099)  120.65
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A=2ax+ay-3az
B=ay-az
C=3ax+5ay+7az
Find:
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(d) A.C - |B|
A.C  ( 2 * 3)  (1* 5)  ( 3 * 7)  6  5  21  10
| B | 1  1 
2
| B |2  2
A.C  | B |2  10  2  12
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A=2ax+ay-3az
B=ay-az
C=3ax+5ay+7az
1
1
1
(d) B x ( A  C )
2
3
4
1
1
1
B
ay az
2
2
2
1
2
1
A
ax 
ay - az
3
3
3
1
3
5
7
C
ax 
ay 
az
4
4
4
4
1
1
17
19
3
(
A
C) 
ax 
ay 
az
3
4
12
12
4
1
1
17
19
3
(
ay az) x (
ax 
ay 
az)
2
2
12
12
4
20
1
1
17
19
3
( ay - az) x ( ax  ay  az)
2
2
12
12
4
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A=5ax+3ay+2az , B=-ax+4ay+6az , C=8ax+2ay
Find α and β such that :
αA+ βB + C is parallel to y-axis
αA+ βB + C= [ 5α ax+3α ay+ 2α az] +
[-β ax+4β ay+6β az] +[8ax+2ay]
=(5α- β+8)ax + (3α +4β +2)ay+ (2α +6β) az
5α- β+8=0 …..(1)
2α +6β=0 ……(2)
solving (1) and (2)
 α=-1.5 , β=-0.5
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A= αax+ 3ay- 2az , B=4ax+ β ay+ 8 az
(a) Find α and β if A an B are parallel?
AxB=|A||B|sin 0 =0
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A= αax+ 3ay- 2az , B=4ax+ β ay+ 8 az
(b) Relationship between α and β if A an B
are perpendicular?
A.B=|A||B| cos 90 =0
4α + 3β -16 = 0
α= 0.25 – 0.75 β
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(b) Show that
ay x az
ax 
ax . ay x az
ay x az
ax



ax . ay x az ax .ax
ax
ax
ax



ax
2
2
| ax || ax | cos 0 | ax |
1
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(3) Scalar Triple Product
A.(BxC)=B.(CxA)= C.(AxB)
= volume of Parallelepiped
‫متوازي السطوح‬
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Show that
A.(BxC)=(AxB).C
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